In the verge of coronavirus pandemic, we are providing FREE access to our entire Online Curriculum to ensure Learning Doesn't STOP!

# Basis Of A Three Dimensional Space

Go back to  'Vectors and 3-D Geometry'

In the preceeding discussion, we talked about the basis of a plane. We can easily extend that discussion to observe that any three non-coplanar vectors can form a basis of three dimensional space:

In other words, any vector $$\vec r$$ in 3-D space can be expressed as a linear combination of three arbitrary non-coplanar vectors. From this, it also follows that for three non-coplanar vectors $$\vec a,\vec b,\vec c,$$ if their linear combination is zero, i.e, if

$\lambda \vec a + \mu \vec b + \gamma \vec c = \vec 0\quad\qquad\qquad\left( {{\text{where}}\,\,\lambda ,\mu ,\gamma \in \mathbb{R}} \right)$

then $$\lambda ,\mu \,\,{\text{and}}\,\,\gamma$$ must all be zero. To prove this, assume the contrary. Then, we have

$\vec a = \left( { - \frac{\mu }{\lambda }} \right)\vec b + \left( { - \frac{\gamma }{\lambda }} \right)\vec c$

which means that $$\vec a$$ can be written as the linear combination of  $$\vec b\,\,{\text{and}}\,\,\vec c$$. However, this would make $$\vec a,\,\,\vec b\,\,{\text{and}}\,\vec c$$ coplanar, contradicting our initial supposition. Thus, $$\lambda ,\mu \,\,{\text{and}}\,\,\gamma$$  must be zero.

We finally come to what we mean by linearly independent and linearly dependent vectors.

Linearly independent vectors : A set of non-zero vectors $${\vec a_1},{\vec a_2},{\vec a_3}....,{\vec a_n}$$  is said to be linearly independent if

${\lambda _1}{\vec a_1} + {\lambda _2}{\vec a_2} + ... + {\lambda _n}{\vec a_n} = \vec 0$

$implies\,\,\,\,{\lambda _1} = {\lambda _2} = .... = {\lambda _n} = 0$

Thus, a linear combination of linearly independent vectors cannot be zero unless all the scalars used to form the linear combination are zero.

Linearly dependent  vectors: A set of non-zero vectors  $${\vec a_1},{\vec a_2},{\vec a_3},....,{\vec a_n}$$  is said to be linearly dependent if there exist scalars $${\lambda _1},{\lambda _2}....{\lambda _n},$$ not all zero such that,

${\lambda _1}{\vec a_1} + {\lambda _2}{\vec a_2} + ..... + {\lambda _n}{\vec a_n} = \vec 0$

For example, based on our previous discussions, we see that

(i) Two non-zero, non-collinear vectors are linearly independent.

(ii) Two collinear vectors are linearly dependent

(iii) Three non-zero, non-coplanar vectors are linearly independent.

(iv) Three coplanar vectors are linearly dependent

(v) Any four vectors in 3-D space are linearly dependent.

You are urged to prove for yourself all these assertions.

Example – 5

Let $${\vec a},\vec b\,\,{\text{and}}\,\,\vec c$$ be non-coplanar vectors. Are the vectors $$2\vec a - \vec b + 3\vec c,\,\,\vec a + \vec b - 2\vec c\,\,{\text{and}}\,\,\vec a + \vec b - 3\vec c$$  coplanar or non-coplanar?

Solution: Three vectors are coplanar if there exist scalars $$\lambda ,\mu \in \mathbb{R}$$ using which one vector can be expressed as the linear combination of the other two.

Let us try to find such scalars:

$2\vec a - \vec b + 3\vec c = \lambda \left( {\vec a + \vec b - 2\vec c} \right) + \mu \left( {\vec a + \vec b - 3\vec c} \right)$

$\Rightarrow \quad \left( {2 - \lambda - \mu } \right)\vec a + \left( { - 1 - \lambda - \mu } \right)\vec b + \left( {3 + 2\lambda + 3\mu } \right)\vec c = \vec 0$

Since $$\vec a,\vec b,\vec c,$$ are non-coplanar, we must have

$2 - \lambda - \mu = 0$

$- 1 - \lambda - \mu = 0$

$3 + 2\lambda + 3\mu = 0$

This system, as can be easily verified , does not have a solution for  $$\lambda \,\,{\text{and}}\,\,\mu$$.

Thus, we cannot find scalars for which one vector can be expressed as the linear combination of the other two, implying the three vectors must be non-coplanar.

As an additional exercise, show that for three non-coplanar vectors $$\vec a,\,\,\vec b\,\,{\text{and}}\,\,\vec c$$ , the vectors  $$\vec a - \,2\vec b\, + 3\vec c,\,\,\vec a - 3\vec b + 5\vec c\,\,\,and\,\,\, - 2\vec a - \,3\vec b\, - 4\vec c$$  are coplanar.

## What is basis and dimension?

• An important result in linear algebra is the following: Every basis for V has the same number of vectors. The number of vectors in a basis for V is called the dimension of V, denoted by dim(V).

## What is a basis of a vector?

• A vector basis of a vector space is defined as a subset of vectors in that are linearly independent and span

## What is the dimensions of a set of vectors?

• Dimension of a Vector Space If V is spanned by a finite set, then V is said to be finite-dimensional, and the dimension of V, written as dim V, is the number of vectors in a basis for V. The dimension of the zero vector space {0} is defined to be 0.

## How to find the angle between two vectors in 3d?

• a. Calculate the magnitude of the vectors.
• b. Divide each vector by its vector magnitude to compute its unit vector.
• c. Compute the dot product of the unit vectors.
• d. Take the arcosine of the dot product of the unit vectors to get the angle between the vectors in radians.

## What is the angle between two vectors?

• The angle between two vectors, deferred by a single point, called the shortest angle at which you have to turn around one of the vectors to the position of co-directional with another vector.
Vectors
grade 11 | Questions Set 1
Vectors
Vectors
grade 11 | Questions Set 2
Vectors
Three Dimensional Geometry
grade 11 | Questions Set 2
Three Dimensional Geometry
Three Dimensional Geometry
grade 11 | Questions Set 1
Three Dimensional Geometry
Vectors
grade 11 | Questions Set 1
Vectors
Vectors
grade 11 | Questions Set 2
Vectors